Australian Math Society Web Site - the Gazette

$i^\wedge i^\wedge i^\wedge i^\wedge \ldots $

R.O. Weber and J. Roumeliotis

We recently wondered about the evaluation of
z = i^\wedge i^\wedge i^\wedge i^\wedge \ldots \tag1
when there are an infinite number of terms. As we shall show, this can be evaluated to give simple answers, albeit an infinite number of them.

One begins by taking the logarithm of both sides of equation (1), yielding
\log z = z\log i. \tag2
(Incidentally, this is reminiscent of Lambert's w-function.) As \log i is multi-valued, this becomes
\frac{1}{z} \log z = (4n +1)\frac{\pi}{2} i. \tag3
Setting z = re^{i\theta}, expanding with Euler's formula for e^{i\theta}, and equating the real and imaginary parts, one finds
\frac{1}{r}(\cos\theta.\log r + (\theta + 2k\pi).\sin\theta) = 0, \tag4

\frac{1}{r}\left( (\theta + 2k\pi).\cos\theta - \sin\theta.\log r \right) = (4n+1)\frac{\pi}{2}. \tag5
Equation (4) furnishes a formulae for r in terms of \theta:
\log r = - (\theta + 2k\pi)\tan\theta. \tag6
Using this once only in equation (5) furnishes a more instructive formula for r in terms of \theta:
r = \frac{2(\theta + 2k\pi)}{(4n+1)\pi\cos\theta}\;, \tag7
while repeated substitution of equation (6) into equation (5) gives an implicit formula for \theta:
\frac{(\theta + 2k\pi)e^{(\theta + 2k\pi)\tan\theta}}{\cos\theta} = (4n+1)\frac{\pi}{2}. \tag8
Clearly for chosen values of k and n, equation (8) can be solved numerically for \theta. The value found can then be used in equation (7), or (6), to find r.

Our particular interest in evaluating i^\wedge i^\wedge i^\wedge i^\wedge \ldots was to search for patterns in the (infinite number of) solutions. For example with k=0, solutions to equation (8) are graphed as shown in figure~ 1. The multiplicity of solutions is evident, as is the asymptotic behaviour of these solutions:
\theta \sim 2m\pi. \tag9
Because of this, \cos\theta \sim 1, whereupon equation (7) gives
r \sim \frac{2}{(4n+1)\pi}\theta. \tag10
That is, on a (r,\theta) diagram, the solutions are asymptotically on a straight line. For other values of k, the solutions are asymptotically upon the straight lines
r = \frac{2}{(4n+1)\pi}(\theta+ 2k\pi). \tag11

To next order, one can solve equation (8) to find
\theta^{(2)} \sim 2m\pi - \frac{1}{2(m + k)\pi} \log\left(\frac{4(m + k)}{4n + 1}\right) \tag12
and equation (11) still holds, but with \theta^{(2)} inserted in place of \theta. Clearly this could be continued to any order. Complementary relationships emerge in the cartesian plane. Setting z = x + iy and substituting (7) into the transformation:
x = r\cos\theta, \qquad\qquad y = r\sin\theta, \tag13
gives:
x = \frac{2(\theta + 2k\pi)}{(4n + 1)\pi} \qquad \text{and} \qquad y = \frac{2(\theta + 2k\pi)}{(4n+1)\pi}\tan\theta. \tag14
Expanding (14) for \theta close to 2m\pi and evaluating one finds:
y \sim \frac{-2}{(4n + 1)\pi} \log x. \tag15
We note that (15) is valid for any k. This note has shown that there are infinitely many values of i^\wedge i^\wedge i^\wedge i^\wedge \ldots and that these values are distributed in the complex plane in a reasonably simple way.
Department of Mathematics
University College, UNSW
ADFA Canberra 2600


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